Diffusion Into a Falling Liquid Film

Diffusion into a Falling Liquid Film We provide a detailed solution to the problem of diffusion into a falling liquid film, following the discussion in BSL1 §18.5. The same problem is treated in WRF2 §26.4 in somewhat less detail. Figure: Sketch of system from BSL. Skipping the exposition, we will start with the governing equation for the system,   x 2  ∂c ∂cA ∂ 2 cA A = vz (x) = vmax 1 − , DAB ∂x2 ∂z δ ∂z along with the boundary conditions ∂cA (x = δ, z) = 0. ∂x The second equality follows from the velocity profile of the flow of a falling film, which was derived last semester. The first two boundary conditions are obvious, whereas the last follows from the vanishing flux in the x-direction at x = δ by virtue of the solid wall there. Note that the boundary data are inconsistent at the point (x, z) = (0, 0), which must therefore be excluded from our analysis. cA (x, z = 0) = 0, cA (x = 0, z) = cA0 , As is, this partial differential equation is quite complicated to solve, so we make a few simplifications. First, replace the boundary condition ∂cA /∂x(x = δ, z) = 0 with cA (x = ∞, z) = 0. This removes the dependence of the solution on δ because there is now no length scale associated with diffusion in the x-direction, so we can solve the governing equations by a clever scaling argument and hence obtain a self-similar solution. We will check later that this replacement does not change the system appreciably in the regime that we consider. The two directions x and z are uncoupled and can be considered independently. In the z-direction, the relevant variables are vz and z, which gives a characteristic time τz ∼ z/vz . In the x-direction, the relevant variables are DAB and x, which gives a characteristic time τx ∼ x2 /DAB . We can roughly interpret τz as the time required to traverse a distance z, and similarly for τx . We will work in the regime τz  τx —i.e., species A does not proceed very far in the x-direction by the time it flows through the system in the z-direction—and consider only x  δ. This allows us to make a second approximation: in this limit,   x 2  ≈ vmax . vz (x) = vmax 1 − δ Finally, we can check for consistency. In the regime τz  τx , we have cA (x ∼ δ, z) ≈ 0 and hence ∂cA /∂x(x = δ, z) ≈ 0, our original boundary condition. The governing equation reduces to DAB ∂ 2 cA ∂cA = vmax , 2 ∂x ∂z along with the boundary conditions cA (x, z = 0) = 0, 1 Bird, cA (x = 0, z) = cA0 , cA (x = ∞, z) = 0. Stewart, and Lightfoot, Transport Phenomena, rev. 2nd ed. Rorrer, and Foster, Fundamentals of Momentum, Heat, and Mass Transfer, 6th ed. 2 Welty, Diffusion into a Falling Liquid Film solution 1. (Self-similar solution.) Make the substitution φ = cA (x, z)/cA0 . φ is a dimensionless function that depends only on the parameters x, z, DAB , and vmax , because these are the only parameters that appear in the governing equation and boundary conditions. Since φ is dimensionless, so too must be its argument. The one physical nondimensional argument3 that we can concoct from the given parameters is η := p x DAB z/vmax , so that φ = φ(η). Plugging this into the governing equation and using the chain rule, we find, after some algebra, that our governing equation reduces to d2 φ η dφ + =0 dη 2 2 dη with boundary conditions φ(η = 0) = 1, φ(η = ∞) = 0. This is promising, because our partial differential equation is now an ordinary differential equation, and can be solved by elementary means. Substitute ϕ := dφ/dη to obtain  2 dϕ η η + ϕ = 0 ⇐⇒ ϕ = C exp − , dη 2 4 where we have integrated the resulting equation directly. A second integration yields  02  Z η Z η η φ=C +D =C dη 0 exp − dη 0 exp(−η 02 ) + D 4 0 0 Z η 2 dη 0 exp(−η 02 ) = 1 − erf(η) = erfc(η). =1− √ π 0 We have changed the variable of integration from η 0 to 2η 0 and hence absorbed a factor of 2 into C and a factor of 1/2 into η in the second equality. The third equality follows from applying boundary conditions. Our final result, in terms of our original variables, is therefore ! x cA (x, z) = cA0 erfc p . 4DAB z/vmax Note that η now has an additional factor of 2 = √ 4 in the denominator because of our redefinition of η. The flux through the interface can then be evaluated as −DAB ∂cA ∂x = −DAB x=0 ∂η ∂cA ∂x ∂η −cA0 DAB η=0 =p 4DAB z/vmax  2 −√ π  ∂ ∂η Z r η 0 02 = cA0 dη exp(−η ) 0 η=0 DAB vmax . πz solution 2. (Separation of variables.) Our governing equation can also be solved by separation of variables. Separation of variables is guaranteed to get all possible solutions here, because the functions of x and z thus obtained are sinusoids and exponentials, each set of which forms a complete basis. Instead of performing the separation of variables process explicitly, we will cheat by starting from our final expression in the first solution and manipulate it into a form that demonstrates a separation of variables. We have, as desired, Z η ∞ ∞ X 2 (−1)n η 02n 2 X (−1)n η 2n+1 0 √ √ dη exp(−η ) = 1 − dη =1− n! π 0 π n=0 n! 2n + 1 0 n=0   ∞ ∞ n+1/2 X 2 X (−1)n 1 vmax =1− √ x2n+1 z −(n+1/2) =: C + an fn (x)gn (z). π n=0 n! 2n + 1 4DAB n=0 2 φ(η) = 1 − √ π Z η 0 02 Further simplification should get us to a form involving sinusoids and exponentials, and this is left as an exercise for the interested reader. 3 Of course any power of η is also nondimensional, but these powers all represent the same physical variable and can be considered the √ same. For mathematical reasons, it is convenient to let the nondimensional argument contain x (as opposed to x, x2 , etc.). And while there are other nondimensional combinations we can come up with—x/z is one example—but these combinations are not physical, because they do not respect the independence of the x- and z-directions. This independence suggests that we should first lump variables within each direction, and then between directions: here, we have η 2 = τx /τz . Diffusion into a Falling Liquid Film solution 3. (Fourier and Laplace transforms.) The integral (Fourier and Laplace, primarily) transforms are particularly useful for solving linear ordinary and partial differential equations, with the Fourier transform in the interval (−∞, ∞) and the Laplace transform in the interval (0− , ∞). The Laplace transform replaces a function φ(x) by Z ∞ φ̂(ξ) = dx φ(x)e−xξ 0 for a generally-complex variable ξ, whereas the Fourier transform replaces a function φ(x) by Z ∞ φ̂(k) = dx φ(x)e−ikx −∞ for a real variable k. The Laplace transform is a special case of the Fourier transform when we allow k to be complex. Both transforms are linear and widely applicable for differential equations because the exponential functions are eigenfunctions under differentiation, and these transforms essentially diagonalize our original function in an exponential-function basis, reducing differentiation to multiplication by the eigenvalues. A whole slew of transform pairs and identities have been derived for the Laplace and Fourier transforms, and are readily available on, say, Wikipedia. Essentially all can be derived from the definition for suitably well-behaved functions. Let us consider the Laplace transform of the governing equation—multiply both sides by the exponential term and integrate from 0 to ∞—with respect to x, yielding ∂ φ̂ = ξ 2 φ̂ − ξφ(0) − φ0 (0). ∂z The second equality follows from an identity of the Laplace transform. We are now stuck, because φ0 (0) is not given. We must therefore transform with respect to z, and this is done in WRF §26.4. We cannot, unfortunately, use the Fourier transform here, because the domain of definition of x (or z) in the problem is not the entire real line R. One might anticipate that we could extend the domain of definition in x by symmetry to the entire real line, but doing so would introduce a cusp at the origin—since the flux at x = 0 is nonzero, and therefore so is ∂cA /∂x there—, which renders our potential solution non-differentiable, and certainly by this point we cannot hope to take two partial derivatives of x. If we symmetrize with respect to z, on the other hand, we return to the Laplace-transformed problem. rf